The DAV Class 8 Maths Book Solutions and **DAV Class 8 Maths Chapter 1 Worksheet 3** Solutions of Squares and Square Roots offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 1 WS 3 Solutions

Question 1.

Find the square root of the following by prime factorisation:

(i) 225

(ii) 441

(iii) 529

(iv) 40000

(v) 7744

(vi) 8281

(vii) 4096

(viii) 28900

Solution:

(i) Resolving 225 into prime factors.

225 = 3 × 3 × 5 × 5

∴ \(\sqrt{225}\) = 3 × 5 = 15

(ii) Resolving 441 into prime factors

441 = 3 × 3 × 7 × 7

∴ \(\sqrt{441}\) = 3 × 7 = 21

(iii) Let us resolve 529 into prime factors.

529 = 23 × 23

∴ \(\sqrt{529}\) = 23

(iv) Let us resolve 40000 into prime factors

40000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

∴ \(\sqrt{40000}\) = 2 × 2 × 2 × 5 × 5 = 200

(v) Resolving 7744 into prime factors

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

∴ \(\sqrt{7744}\) = 2 × 2 × 2 × 11 = 88

(vi) Resolving 8281 into prime factors

8281 = 7 × 7 × 13 × 13

∴ \(\sqrt{8281}\) = 7 × 13 = 91

(vii) Resolving 4096 into prime factors

4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ \(\sqrt{4096}\) = 2 × 2 × 2 × 2 × 2× 2 = 64

(viii) Resolving 28900 into prime factors

28900 = 2 × 2 × 5 × 5 × 17 × 17

∴ \(\sqrt{28900}\) = 2 × 5 × 17 = 170.

Question 2.

Find the smallest number by which 1100 must be multiplied so that the product becomes a perfect square. Also, find the square root of the perfect square so obtained.

Solution:

Resolving 1100 into prime factors

1100 = 2 × 2 × 5 × 5 × 11

Here, 11 is unpaired. So 11 must be multiplied to 1100 to obtain a perfect square number.

∴ The perfect square number so obtained = 1100 × 11 = 12100

Question 3.

By what smallest number must 180 be multiplied so that it becomes a perfect square?

Also find the square root of the number so obtained.

Solution:

Resolving 180 into prime factors.

180 = 2 × 2 × 3 × 3 × 5

Here 5 is unpaired.

∴ 5 is required to multiply 180 to make it perfect square.

∴ Required number is 5 × 180 = 900

∴ \(\sqrt{900}\) = 2 × 2 × 3 × 3 × 5 × 5

= 2 × 3 × 5 = 30

Question 4.

Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.

Solution:

Resolving 3645 into prime factors.

3645 = 3 × 3 × 3 × 3 × 3 × 3 × 5

Here 5 is unpaired. So the given number 3645 must be divided by 5 to obtain a perfect square.

So the required number = 3645 ÷ 5 = 729

∴ \(\sqrt{729}\) = 3 × 3 × 3 × 3 × 3 × 3

= 3 × 3 × 3 = 27

Question 5.

A gardener planted 1521 trees in rows such that the number of rows was equal to the number of plants in each row. Find the number of rows.

Solution:

Let the number of rows be x.

number of plants in each row = x

Total number of plants = x × x = x^{2}

∴ x^{2} = 1521

⇒ x = \(\sqrt{1521}\)

⇒ 1521 = 3 × 3 × 13 × 13

⇒ \(\sqrt{1521}\) = 3 × 13 = 39

Hence the number of rows is 39.

Question 6.

An officer wants to arrange 202500 cadets in the form of a square. How many cadets were there in each row?

Solution:

Let there be x number of cadets in each row.

∴ Number of rows = x

Total number of cadets = x × x = x^{2}

x^{2} = 202500

x = \(\sqrt{202500}\)

202500 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 5

∴ \(\sqrt{202500}\) = 2 × 3 × 3 × 5 × 5 = 450

Hence, the number of cadets in each row = 450.

Question 7.

The area of a square field is 5184 m^{2}. A rectangular field whose length is twice its breadth, has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.

Solution:

The area of the square field = 5184 m^{2}

∴ side of the square = \(\sqrt{5184}\) m

Resolving 5184 into prime factors

5184 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

∴ \(\sqrt{5184}\) = 2 × 2 × 2 × 3 × 3 = 72

∴ Perimeter of the square field = 4 × side

= 4 × 72 = 288 m

Now, let the breadth of the rectangular field be x m.

∴ Length = 2x m

Perimeter of the rectangular field = 2 [l + b] = 2 [2x + x] = 6x m

∴ 6x = 288

⇒ x = \(\frac {288}{6}\)

⇒ x = 48 m.

∴ breadth = 48 m.

and length = 2 × 48 = 96 m

∴ Area of the rectangular field = l × b = 96 × 48 = 4608 m^{2}

Hence the required area = 4608 m^{2}

Question 8.

Find the value of \(\sqrt{47089}+\sqrt{24336}\)

Solution:

Resolving 47089 into prime factors,

47089 = 7 × 7 × 31 × 31

∴ \(\sqrt{47089}\) = 7 × 31 = 217

Resolving 24336 into prime factors,

24336 = 2 × 2 × 2 × 2 × 3 × 3 × 13 × 13

∴ \(\sqrt{24336}\) = 2 × 2 × 3 × 13 = 156

Thus, \(\sqrt{47089}+\sqrt{24336}\) = 217 + 156 = 373.

Additional Questions:

Question 1.

Find the square root of each of the following numbers by prime factorisation.

(i) 130321

(ii) 46656

(iii) 45796

Solution:

(i) Resolving 130321 into prime factors,

∴ 130321 = 19 × 19 × 19

7130321 = 19 × 19 = 361

(ii) Resolving 46656 into prime factors,

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

746656 = 2 × 2 × 2 × 3 × 3 × 3 = 216

(iii) Resolving 45796 into prime factors,

∴ 45796 = 2 × 2 × 107 × 107

\(\sqrt{45796}\) = 2 × 107 = 214.

Question 2.

Is 2352 a perfect square? If not, find the smallest multiple of 2352 which is a perfect square. Find the square root of the number so obtained.

Solution:

Resolving 2352 into prime factors

∴ 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7

Here 3 is unpaired. So the given number 2352 is not a perfect square.

∴ 3 is needed to make the given number 2352 as a perfect square.

Hence the smallest multiple of 2352 is 3 which makes it a perfect square.

∴ Perfect square number = 2352 × 3 = 7056

7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

\(\sqrt{7056}\) = 2 × 2 × 3 × 7 = 84

Question 3.

Find the smallest number by which 1008 must be multiplied to make it 2 252 a perfect square. 2 126

Solution:

Resolving 1008 into prime factors,

∴ 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Hence, 7 is unpaired. So 7 is required to make a pair of 7 which is the 7 smallest number to make the given number 1008 a perfect square. Hence, 7 is the required number.

Question 4.

Find the smallest number by which 2645 must be divided so that it 23 529 becomes a perfect square. Also, find the square root of the resulting 23 23 number.

Solution:

Resolving 2645 into prime factors

∴ 2645 = 5 × 23 × 23

Here 5 has no pair. So the given number 2645 must be divided by 5 to make it a perfect square.

∴ Resulting number = 2645 ÷ 5 = 529

∴ \(\sqrt{529}\) = 23

Question 5.

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Solution:

Let number of rows be x

∴ Number of plants in each row = x

Total number of plants = x × x = x^{2}

∴ x^{2} = 2025

⇒ x = \(\sqrt{2025}\)

Resolving 2025 into prime factors,

2025 = 3 × 3 × 3 × 3 × 5 × 5

\(\sqrt{2025}\) = 3 × 3 × 5 = 45

Hence the number of plants in each row = 45

and number of rows = 45

Question 6.

Find the smallest square number which is divisible by each of the numbers 6, 9 and 15.

Solution:

The smallest number divisible by 6, 9 and 15 is their L.C.M. which is equal to 2 × 3 × 3 × 5 = 90

Here prime factors are 90 = 2 × 3 × 3 × 5

2 and 5 are unpaired. So 90 is not a perfect square.

∴ 2 × 5 = 10 is needed to multiply 90 to make it a prefect square.

The required smallest number = 90 × 10 = 900.

Question 7.

Find the length of the side of a square whose area = 1024 m^{2}.

Solution:

Area of a square of side x m = x^{2} m^{2}

∴ x^{2} = 1024

⇒ x = \(\sqrt{1024}\)

Resolving 1024 into prefect square,

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ \(\sqrt{1024}\) = 2 × 2 × 2 × 2 × 2 = 32

Hence, the length of the side of a square = 32 m.

Question 8.

The area of a square field is 256 m^{2} whose perimeter is equal to the perimeter of a rectangular field whose length is thrice its breadth. Find the dimensions of the rectangular field.

Solution:

Area of the square field = 256 m^{2}

∴ Side of the square field = \(\sqrt{256}\)

Resolving 256 into prime factors,

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ \(\sqrt{256}\) = 2 × 2 × 2 × 2 = 16

Side of the square field = 16 m

∴ Its perimeter = 4 × 16 = 64 m.

Now, let x m be the breadth of the rectangular field.

∴ Its length = 3x m

Perimeter = 2[l + b] = 2[3x + x] = 8x m

∴ 8x = 64 ⇒ x = 8 m.

Hence the breadth = 8 m and length = 3 × 8 = 24 m.

Question 9.

In a right ∆ABC, ∠B = 90°. If AB = 20 cm, BC = 21 cm, find AC.

Solution:

From Pythagoras Theorem,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (20)^{2} + (21)^{2}

= 400 + 441 = 841

∴ AC = \(\sqrt{841}\)

Now resolving 841 into prime factors,

841 = 29 × 29

∴ \(\sqrt{841}\) = 29

Hence AC = 29 m.

Question 10.

There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?

Solution:

Let number of rows be x.

∴ number of columns = x

Total number of students = x × x = x^{2}

∴ x^{2} = 500

x = \(\sqrt{500}\)

Resolving 500 into prime factors

500 = 2 × 2 × 5 × 5 × 5

∴ \(\sqrt{500}\) = 2 × 5

Here 5 is unpaired. So, 5 students would be left out in this arrangement.